JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(600\,mL\) of \(0.01\,M\,HCl\) is mixed with \(400\,mL\) of \(0.01\,M\,H _2 SO _4\). The \(pH\) of the mixture is \(.......\times 10^{-2}\). (Nearest integer) [Given \(\log 2=0.30, \quad \log 3=0.48\) \(\log 5=0.69 \quad \log 7=0.84\) \(\log 11=1.04]\)
- A \(185\)
- B \(186\)
- C \(184\)
- D \(183\)
Answer & Solution
Correct Answer
(B) \(186\)
Step-by-step Solution
Detailed explanation
Total milimoles of \(H ^{+}=(600 \times 0.01)+(400 \times 0.01 \times 2)\) \(=14\) \(\left[ H ^{+}\right]=\frac{14}{1000}=14 \times 10^{-3}\) \(pH =3-\log 14\) \(=1.86\) \(=186 \times 10^{-2}\)
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